Introduction

In this blog post, we will explore how to solve the LeetCode problem "2304" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given a 0-indexed m x n integer matrix grid consisting of distinct integers from 0 to m n - 1. You can move in this matrix from a cell to any other cell in the next row. That is, if you are in cell (x, y) such that x < m - 1, you can move to any of the cells (x + 1, 0), (x + 1, 1), ..., (x + 1, n - 1). Note that it is not possible to move from cells in the last row. Each possible move has a cost given by a 0-indexed 2D array moveCost of size (m n) x n, where moveCost[i][j] is the cost of moving from a cell with value i to a cell in column j of the next row. The cost of moving from cells in the last row of grid can be ignored. The cost of a path in grid is the sum of all values of cells visited plus the sum of costs of all the moves made. Return the minimum cost of a path that starts from any cell in the first row and ends at any cell in the last row. Example 1: Input: grid = [[5,3],[4,0],[2,1]], moveCost = [[9,8],[1,5],[10,12],[18,6],[2,4],[14,3]] Output: 17 Explanation: The path with the minimum possible cost is the path 5 -> 0 -> 1. - The sum of the values of cells visited is 5 + 0 + 1 = 6. - The cost of moving from 5 to 0 is 3. - The cost of moving from 0 to 1 is 8. So the total cost of the path is 6 + 3 + 8 = 17. Example 2: Input: grid = [[5,1,2],[4,0,3]], moveCost = [[12,10,15],[20,23,8],[21,7,1],[8,1,13],[9,10,25],[5,3,2]] Output: 6 Explanation: The path with the minimum possible cost is the path 2 -> 3. - The sum of the values of cells visited is 2 + 3 = 5. - The cost of moving from 2 to 3 is 1. So the total cost of this path is 5 + 1 = 6. Constraints: m == grid.length n == grid[i].length 2 <= m, n <= 50 grid consists of distinct integers from 0 to m n - 1. moveCost.length == m n moveCost[i].length == n 1 <= moveCost[i][j] <= 100

Explanation

Here's the breakdown of the solution:

• Dynamic Programming: Use dynamic programming to store the minimum cost to reach each cell in the grid. The state dp[i][j] represents the minimum cost to reach cell (i, j).
• Iteration: Iterate through the rows of the grid, updating the dp array based on the minimum cost to reach cells in the previous row and the move costs.

• Base Case: Initialize the first row of the dp array with the values from the first row of the grid.

• Runtime Complexity: O(m*n^2)

• Storage Complexity: O(m*n)

Code

    def minCost(grid, moveCost):
    m = len(grid)
    n = len(grid[0])
    dp = [[0] * n for _ in range(m)]

    # Initialize the first row of dp with the values from the first row of grid
    for j in range(n):
        dp[0][j] = grid[0][j]

    # Iterate through the rows of the grid, starting from the second row
    for i in range(1, m):
        for j in range(n):
            dp[i][j] = float('inf')
            # Iterate through the columns of the previous row
            for k in range(n):
                # Calculate the cost of moving from (i-1, k) to (i, j)
                cost = dp[i - 1][k] + moveCost[grid[i - 1][k]][j] + grid[i][j]
                # Update dp[i][j] with the minimum cost
                dp[i][j] = min(dp[i][j], cost)

    # Find the minimum cost to reach any cell in the last row
    min_cost = float('inf')
    for j in range(n):
        min_cost = min(min_cost, dp[m - 1][j])

    return min_cost